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A certain flexible weather balloon contains helium gas at a volume of 855 L. Initially, the balloon is at sea level where the temperature is \(25^{\circ} \mathrm{C}\) and the barometric pressure is 730 torr. The balloon then rises to an altitude of 6000 ft, where the pressure is 605 torr and the temperature is \(15^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from sea level to 6000 ft?

Short Answer

Expert verified
The change in volume of the weather balloon as it ascends from sea level to 6000 ft is 167.32 L.

Step by step solution

01

Convert temperatures into Kelvin

To work with the gas laws, we should convert the given temperatures in Celsius to Kelvin. To do this, we add 273.15 to the Celsius temperatures: T1 = \(25^{\circ} \mathrm{C} + 273.15 = 298.15\ \mathrm{K}\) T2 = \(15^{\circ} \mathrm{C} + 273.15 = 288.15\ \mathrm{K}\)
02

Convert pressures into atmospheres

To simplify our calculations, we should convert the given pressures from torr to atmospheres (atm). To do this, we divide the torr values by 760: P1 = 730 torr / 760 = 0.9605 atm P2 = 605 torr / 760 = 0.7961 atm
03

Setup the combined gas law equation

Now, we have T1, T2, P1, and P2 in the appropriate units. We can now use the combined gas law equation, remembering that V1 = 855 L: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\)
04

Solve the equation for the final volume (V2)

Now we need to solve for V2. We can do this by rearranging the combined gas law equation and plugging in the known values: \(V_2 = \frac{P_1V_1T_2}{P_2T_1}\) Plug in all values: \(V_2 = \frac{(0.9605\ \mathrm{atm})(855\ \mathrm{L})(288.15\ \mathrm{K})}{(0.7961\ \mathrm{atm})(298.15\ \mathrm{K})}\)
05

Calculate the final volume (V2)

Perform the above calculation to find V2: \(V_2 = 1022.32\ \mathrm{L}\)
06

Calculate the change in volume

The change in volume is the final volume minus the initial volume: 螖V = V2 - V1 = 1022.32 L - 855 L = 167.32 L As the balloon ascends from sea level to 6000 ft, its volume increases by 167.32 L.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The Ideal Gas Law is a fundamental equation that describes the behavior of gases. It relates the four important properties of a gas: pressure (\(P\)), volume (\(V\)), temperature (\(T\)), and the amount of gas, often represented as the number of moles (\(n\)). The ideal gas law formula is given by:\[PV = nRT\]where \(R\) is the universal gas constant.
The Ideal Gas Law helps us predict how a gas will behave under different conditions, such as changing temperature or pressure. In many practical situations, like in this problem involving a helium balloon, we assume that gases behave ideally, which grants us a simpler way to estimate changes in a gas' properties.
In scenarios where the number of moles does not change, and instead only the volume, temperature, and pressure are altered, the Combined Gas Law comes into play, maintaining the core concepts of the Ideal Gas Law.
Volume and Pressure Relationship
The relationship between pressure and volume of a gas is described by Boyle's Law, which states that at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure. In simpler terms, as the pressure on a gas increases, its volume decreases, and vice versa.
  • This can be mathematically expressed as: \[PV = ext{constant}\]
  • When using the Combined Gas Law in our problem, we see that pressure and volume are linked through \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\), meaning changes in pressure and temperature will affect volume.
When solving gas-related problems, understanding this relationship helps us manipulate and predict the final volume a gas occupies when pressure conditions change, such as when a weather balloon rises to a higher altitude.
Temperature and Gas Behavior
The behavior of gas molecules is significantly influenced by changes in temperature. When temperature increases, the kinetic energy of gas molecules also increases, causing them to move more vigorously, which can lead to an increase in the volume of the gas if the pressure is constant.
For calculations involving gases, it鈥檚 crucial to use Kelvin (\(K\)), the absolute temperature scale. The conversion from Celsius to Kelvin is straightforward: just add 273.15 to the Celsius temperature.
  • For instance, in our balloon problem, we converted 25掳C to 298.15 K.
  • Temperature changes are directly handled in the Combined Gas Law: \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\).
Recognizing how temperature influences volume and pressure through kinetic energy allows us to predict changes in gaseous systems effectively.
Helium Gas Properties
Helium is a noble gas known for its light weight and non-reactive nature. These properties make it an ideal choice for filling balloons, especially weather and scientific balloons. When helium is used in a weather balloon, like in this exercise, it aids in illustrating a larger volume expansion compared to heavier gases.
  • Unlike heavier gases, helium rises in the atmosphere because it is less dense than the surrounding air.
  • This lower density helps with reaching higher altitudes, making helium balloons useful for high-altitude observations.
  • Moreover, as a monatomic gas, helium exhibits ideal gas behavior more closely compared to polyatomic gases!
Thus, understanding helium's properties helps in comprehending why it demonstrates significant volume changes under differing atmospheric conditions.

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Most popular questions from this chapter

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g),\) ammonia, \(\mathrm{NH}_{3}(g),\) and oxygen, \(\mathrm{O}_{2}(g),\) at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of \(\mathrm{HCN}(g)\) can be obtained from the reaction of \(20.0 \mathrm{L} \mathrm{CH}_{4}(g), 20.0 \mathrm{L} \mathrm{NH}_{3}(g),\) and 20.0 \(\mathrm{L} \mathrm{O}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.

Consider separate \(1.0-\mathrm{L}\) samples of \(\mathrm{He}(g)\) and \(\mathrm{UF}_{6}(g),\) both at 1.00 atm and containing the same number of moles. What ratio of temperatures for the two samples would produce the same root mean square velocity?

Silane, SiH, , is the silicon analogue of methane, \(\mathrm{CH}_{4}\) . It is prepared industrially according to the following equations: $$\begin{array}{c}{\mathrm{Si}(s)+3 \mathrm{HCl}(g) \longrightarrow \mathrm{HSiCl}_{3}(l)+\mathrm{H}_{2}(g)} \\ {4 \mathrm{HSiCl}_{3}(l) \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l)}\end{array}$$ a. If \(156 \mathrm{mL}\) \(\mathrm{HSiCl}_{3} (d=1.34 \mathrm{g} / \mathrm{mL})\) is isolated when 15.0 \(\mathrm{L}\) \(\mathrm{HCl}\) at 10.0 \(\mathrm{atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3} ?\) b. When \(156 \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at 10.0 \(\mathrm{atm}\) and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is 93.1\(\% ?\)

You have a helium balloon at 1.00 atm and \(25^{\circ} \mathrm{C} .\) You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is 79.0\(\%\) nitrogen and 21.0\(\%\) oxygen by volume. The 鈥渓ift鈥 of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C} ?\) Explain. b. Calculate the temperature of the air required for the hot-air balloon to provide the same lift as the helium balloon at 1.00 atm and \(25^{\circ} \mathrm{C}\) . Assume atmospheric conditions are 1.00 atm and \(25^{\circ} \mathrm{C} .\)

Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at 0.959 atm and 298 \(\mathrm{K}\). Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at 1.51 atm and 375 \(\mathrm{K}\). This mixture has a density of 1.391 g/L and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon

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